Question

If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant?

Answer 1

Answer:

The correct answer is option B.

Explanation:

$3NaOH+2NBr_3\rightarrow 3HOBr+3NaBr+N_2$

Moles of $NBr_3$ = 40 mol

Moles of NaOH = 48 mol

According to reaction, 3 moles of NaOH reacts with 2 moles $NBr_3$

Then ,48 moles of NaOH will reacts with:

$\frac{2}{3}\times 48 mol=32 mol$ of $NBr_3$

Then ,40 moles of $NaBr_3$ will reacts with:

$\frac{3}{2}\times 40 mol=60 mol$ of NaOH

As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.

Moles left after reaction = 40 mol - 32 mol = 8 mol

Hence, the $NBr_3$ is an excessive reagent.

Answer 2

Answer : The correct option is (B) $NBr_3$

Solution : Given,

Moles of $NBr_3$ = 40 mol

Moles of $NaOH$ = 48 mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2NBr_3+3NaOH\rightarrow N_2+3NaBr+3HOBr$

From the balanced reaction we conclude that

As, 3 mole of $NaOH$ react with 2 mole of $NBr_3$

So, 48 moles of $NaOH$ react with $\frac{48}{3}\times 2=32$ moles of $NBr_3$

From this we conclude that, $NBr_3$ is an excess reagent because the given moles are greater than the required moles and $NaOH$ is a limiting reagent and it limits the formation of product.

Excess moles of $NBr_3$ = 40 - 32 = 8 moles

Hence, the correct option is (B) $NBr_3$